Problem 38 - College Algebra and Calculus: An Applied Approach

Question

Find a polynomial with real coefficients that has the given zeros. (There are many correct answers.) $$6,-5+2 i,-5-2 i$$

Step-by-Step Solution

Answer

The polynomial with real coefficients for the given roots \(6, -5+2i, -5-2i\) is \(x^3 +4x^2 -39x + 126\).

Step 1 - Write down the Factors

Given the roots \(6, -5+2i, -5-2i\). Remember the factorized form of the polynomial would look like \( (x - a)(x - b)(x - c) \) . Substituting the given roots we have \( (x - 6)(x -(-5 +2i))(x -(-5 -2i)) \).

Step 2 - Simplify the Factors

Simplify the factors to \( (x - 6)(x + 5 -2i)(x + 5 +2i) \)

Step 3 - Expand the Factors with Complex Numbers

Compute the product \( (x + 5 - 2i)(x + 5 + 2i) \) applying the formula \( (a - b)(a + b) = a^2 - b^2 \), we have \( (x+5)^2 - (2i)^2 = x^2 +10x +25 -4 = x^2 + 10x + 21 \).

Step 4 - Expand all the factors

Now, expand all the factors \( (x - 6)(x^2 + 10x +21) \). After simplifying, we obtain the polynomial \(x^3 +4x^2 -39x + 126 \).

Key Concepts

Complex Zeros

When studying polynomials, it's essential to understand that a zero of a polynomial is a solution to the equation obtained by setting the polynomial equal to zero. But what happens when these solutions are not real numbers? Polynomials with real coefficients can also have complex zeros, which are numbers that have both a real part and an imaginary part. An important concept to grasp is that complex zeros in polynomials with real coefficients always come in pairs, known as complex conjugates. For example, if \( -5+2i \) is a zero, then its complex conjugate \( -5-2i \) is also a zero. This pairing is a direct consequence of the Fundamental Theorem of Algebra and the complex conjugate root theorem, ensuring that the polynomial will have coefficients that are all real.

Expanding Polynomials

Expanding a polynomial is a way of expressing it without parentheses, by performing multiplication between the terms. It’s a fundamental process in algebra that one often encounters when dealing with polynomials. For example, when given the roots of a polynomial, to find the explicit form of the polynomial, you'd need to multiply out the factors corresponding to each root. This means applying distributive properties such as the FOIL (First, Outer, Inner, Last) method or using special product formulas like \(a + b)^2 = a^2 + 2ab + b^2\) and \(a - b)(a + b) = a^2 - b^2\). In the case of the given exercise, expanding the binomials with the complex zeros was simplified using the difference of squares formula, which is critical for handling terms involving complex numbers.

Factoring Polynomials

Factoring polynomials is the reverse process of expanding polynomials. It involves breaking down a complex expression into simpler ones that, when multiplied together, will give the original polynomial. This is particularly useful when solving polynomial equations or simplifying algebraic expressions. Each factor obtained corresponds to a root of the polynomial. With real numbers, factoring can often be done by looking for common patterns or by employing the quadratic formula for second-degree polynomials. However, when dealing with complex zeros, it is helpful to know the zeros ahead as is shown in our exercise, which simplifies the process of factoring the polynomial considerably.

Complex Conjugate Roots

As mentioned earlier, the complex conjugate roots theorem states that if a polynomial has real coefficients and a complex number as a root, then its complex conjugate is also a root. This theorem is crucial when we are dealing with polynomials that have complex zeros. A complex number and its conjugate have the form \(a + bi\) and \(a - bi\), respectively, where \(a\) is the real part and \(bi\) is the imaginary part of the complex number. In the context of the exercise, the roots \( -5+2i \) and \( -5-2i \) are complex conjugates, ensuring that when multiplied, they yield a polynomial with real coefficients, such as \(x^2 + 10x + 21\). This is because the product of conjugates results in the sum of two squares, here \(25 + 4\), thus eliminating the imaginary components and maintaining real coefficients in the polynomial.